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Ma and Pa and brother and me. The sum of our ages is eighty-three. Six times Pa’s age is seven times Ma’s age, and Ma’s age is three times my age.

What is Pa’s age? What is Ma’s age? What is my brother’s age? What is my age?

I try setting up the first equation in four variables letting $A$ be ma's age, $B$ be pa's age, $C$ be my brother's age and $D$ be my age.

$$A+B+C+D=83$$

The problem gives

\begin{align} 7B&=6A\\ A&=3D \end{align}

but I am still lost observing the fact that there was not any fact on my brother.

Martin Thoma
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Acade
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2 Answers2

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If Pa’s age $=7a,$ Ma’s age $=6a,$ your age$=2a$

If your brother’s age $=b,$ we have $7a+6a+2a+b=83\iff b=83-15a$

If ages are integers, $a\le5$

If $a=5,b=8$

If $a=4,b=23$ and Ma’s age $=6a=24$ which is impossible

lab bhattacharjee
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You got the first equation backwards. It should be $6B=7A$. (That is, six times Pa's age, which you've called $B$, is seven times Ma's age, which you've called $A$.) So we have

$$\begin{align} A+B+C+D&=83\\ 6B&=7A\\ A&=3D\\ \end{align}$$

Can you take it from here? (Hint: Ages are typically positive whole numbers, and parents should be a good deal older than their children.)

Barry Cipra
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  • this was what i got and i got stock after i came up with an equation that had two variables i still need help – Acade Nov 02 '14 at 21:36
  • @Acade, does lab bhattacharjee's answer make more sense? You are quite right that, as a pure math problem, three linear equations are not enough to specify four unknowns, but as a "puzzle" problem, we can bring some additional common sense to it. – Barry Cipra Nov 02 '14 at 23:14